Document Sample

Module 3 Homework Chapter 6 – 6.50 (omit normal probability plot), 6.51abcde (S&P only will be fine) Chapter 7 – 7.32, 7.49 Additional Homework 1. Assume that you own a telephone call center. Over the past several years you have determined the following information about the length of telephone call you receive. Your data set of call lengths is normally distributed: Call length, µ = 150 seconds 2 σ = 225 seconds squared a. What is the probability that a phone call will last between 150 and 180 seconds? b. What is the probability that a phone call will last more than 180 seconds? c. What is the probability that a phone call will last between 125 and 150 seconds? d. What is the probability that a phone call will last less than 125 seconds? e. What is the probability that a phone call will last between 145 and 155 seconds? f. What is the probability that a phone call will last between 140 and 155 seconds? g. What is the probability that a phone call will last between 160 and 165 seconds? h. 15% of the phone calls will last less than how many seconds? 2. A manufacturer of tires makes tires with an average life of 36,500 miles (then the tires are worn out). If the standard deviation is 5,000 miles and the distribution of tire life’s is normally distributed… a. What percentage of tires will equal to or exceed 40,000 miles? b. To provide a warranty such that no more than 10% of the tires would have to be replaced, at what mileage should the manufacturer set the warranty? Page 1 of 7 3. Companies that sell stock mutual funds charge their investors expense fees to offset the costs of research and administrative activities. The Funds20153 spreadsheet shows mutual fund fees for 137 funds, expressed as a ratio of total assets invested in the fund. Assume that the distribution of expense ratios is distributed as a normal random variable with the 137 funds the entire population. a. Sort the funds by expense ratio. i. Data, Sort, Expand the Selection, sort ascending by exp. ratio b. Find the mean and standard deviation via Excel. c. Create a frequency distribution histogram and graph it. i. Use a class width of .2, the first class should be .2<=X<.4 d. Find the probability that a stock mutual fund has an expense fee ratio less than 1.5. First, use the Excel function to get the Z value, then use the Excel function to get the probability (area under the curve). e. Find the probability that a mutual fund has an expense fee ration less than .7 f. Find the probability that a mutual fund has an expense fee between 1.4 and 2.0. Excel Functions: =Average( range ) =Stdevp( range ) =Standardize( X, mean, std. dev.) returns the Z value =Normsdist(Z) returns area under curve to the left Page 2 of 7 4. The following data represent the number of days absent per year in a population of 6 employees of a small company: 1 3 6 7 7 12 a. compute the population mean and standard deviation b. select all possible samples of size 2 and set up the sampling distribution of the mean c. compute the mean of the sample means, compute the standard error d. select all possible samples of size 3 and set up the sampling distribution of the mean e. compute the mean of the sample means, compute the standard error f. compare the shape of sampling distributions for b. and d. 5. The diameter of Ping-Pong balls manufactured at a large factory is expected to be approximately normally distributed with a mean of 1.3 inches and a standard deviation of .04 inches. What is the probability that a randomly selected Ping-Pong ball will have a diameter a. Less than 1.28 inches? b. Between 1.28 and 1.32 inches? c. Between what two values (symmetrically distributed around the mean) will 60% of the Ping-Pong balls fall (in terms of diameter)? If many random samples of 16 Ping-Pong balls are selected d. What will be the values of the population mean and standard error of the mean? e. What distribution will the sample means follow? f. What proportion of the sample means will be less than 1.28 inches? g. What proportion of the sample means will be between 1.28 and 1.32 inches? h. Between what two values (symmetrically distributed around the mean) will 60% of the sample means be? i. Which is more likely to occur – an individual ball between 1.28 and 1.32 inches or a sample mean between 1.28 and 1.32 inches? Explain. 6. A call center has a history of receiving calls that last an average of 150 seconds with a standard deviation of 15 seconds. a. What is the probability that a single call is between 150 and 155 seconds? b. What is the probability that a sample of 35 calls will yield a mean between 150 and 155 seconds? Page 3 of 7 Module 3 Homework Answers 6.50 Waiting times resemble an exponential distribution, seating times more closely resemble a normal distribution. Page 4 of 7 6.51 S&P a P(X>0)=P(Z>1.925) 0.0271 b P(X>10)=P(Z>2.425) 0.0077 c P(X<-50)=P(Z<-0.575) 0.2826 d P(X<-60)=P(Z<-1.075) 0.1412 7.32 a. P(p<.5)=P(Z<-1.2087)=0.1134 b. P(p<.5)=P(Z<-2.7028)=0.0034 c. The standard error is decreased. The sampling distribution of the proportion becomes more concentrated around the true proportion of 0.56 and, hence, the probability in b. becomes smaller than in a. 7.49 (a) 24.03 = 10 P(X < 0) = P(Z < -2.403) = 0.0081 (b) P(10 < X < 20) = P(-1.403 < Z < -0.403) = 0.2632 7.49 (c) P(X> 10) = P(Z > -1.403) = 0.9197 10 (d) X = 24.03, X = =5 n 4 P( X < 0) = P(Z < -4.806) = 0.0000 (e) P(10 < X < 20) = P(-2.806 < Z < -0.806) = 0.2076 (f) P( X > 10) = P(Z < -2.806) = 0.9975 (g) Since the sample mean of return of a sample of treasury bond is distributed closer than the return of a single treasury bond to the population mean, the probabilities in (a) and (b) are greater than those in (d) and (e) while the probability in (c) is smaller than that in (f). Page 5 of 7 Additional Homework Answers 1. µ = 150 seconds, σ = 15 seconds a. .4772 b. .0228 c. .4525 d. .0475 e. .2586 f. .3779 g. .0927 or 9.27% h. 134.5 seconds 2. µ = 36,500 miles, σ = 5,000 miles a. .242 or 24.2% b. 30,100 miles 3. Fund Exp Ratio X 1.5000 VANGUARD Sm Gr 0.24 Z 0.3659 FIDELITY SPARTAN Mkt Ind 0.27 Mean 1.34 P(X<1.5) 0.6428 AMER CENT Target 2020 0.59 Std Dev Pop 0.43 PRNCIPL PRESV PSE Tech 0.60 X 0.7000 MAS Mid Cp Gr Inst 0.62 f Z -1.5002 LORD ABBETT Affiliated 0.63 .2<=X<.4 2 P(X<.7) 0.0668 SALOMON Inv Value 0.63 .4<=X<.6 1 VANGUARD Select Val 0.65 .6<=X<.8 11 X 1.4000 AMER CENT Gl Gold 0.69 .8<=X<1.0 13 Z 0.1326 USAA Aggressive Gr 0.72 1.0<=X<1.2 24 P(X<1.4) 0.5528 VANGUARD Explorer Fund 0.74 1.2<=X<1.4 25 PRICE Div Gr 0.77 1.4<=X<1.6 23 X 2.0000 GUARDIAN Park Ave A 0.78 1.6<=X<1.8 17 Z 1.5322 BABSON Growth 0.79 1.8<=X<2.0 10 P(X<2) 0.9373 AMER EXPRESS Stock 0.82 2.0<=X<2.2 8 AMER EXPRESS Mutual 0.83 2.2<=X<2.4 1 P(X1.4<X<2) 0.3845 Fremont Global 0.85 2.4<=X<2.6 2 PIMCO Small cap value Inst 0.85 137 SMITH BREEDEN Eq + 0.88 JANUS Worldwide 0.89 AMER AADVANT Int Eq 0.90 KEMPER Tech 0.93 3-Yr. Returns freq. distribution FRANKLIN Small Cap Gr A 0.94 Number of Returns JP MORGAN Sm Co 0.97 30 JANUS Enterprise 0.98 25 MFS Research A 0.98 20 PUTNAM OTC Emerg Gr 0.98 15 AMER CENT Giftrust 1.00 10 AMER CENT Vista 1.00 5 BRINSON Global equity 1.00 0 FRANKLIN Dynatech 1.00 .4 .6 <= 8 .2 .4 .6 .8 .0 .2 .4 .6 0 . 1. NORTHERN Sm Cp 1.00 <1 <1 <1 <1 <2 <2 <2 <2 X< X< X< X< =X =X =X =X =X =X =X =X <= <= <= KEMPER Samall cap equity A 1.01 0< 2< 4< 6< 8< 0< 2< 4< .2 .4 .6 .8 FPA Paramount 1.03 1. 1. 1. 1. 1. 2. 2. 2. AMER EXPRESS Discovery 1.05 Expense Ratio FLAG Communications 1.05 FRANKLIN Global Comm A 1.05 HOTCHKISS and WILEY Small Cap 1.05 Page 6 of 7 4. X X Sample Mean X Sample Mean 1 1,3 2 1,3,6 3.33 3 1,6 3.5 1,3,7 3.67 6 1,7 4 1,3,7 3.67 7 1,7 4 1,3,12 5.33 7 1,12 6.5 1,6,7 4.67 12 3,6 4.5 1,6,7 4.67 3,7 5 1,6,12 6.33 Mean 6 3,7 5 3,6,7 5.33 Stdev 3.46 3,12 7.5 3,6,7 5.33 6,7 6.5 3,6,12 7 6 C2 15 6,7 6.5 6,7,7 6.67 6 C3 20 6,12 9 6,7,12 8.33 7,7 7 6,7,12 8.33 7,12 9.5 7,7,12 8.67 7,12 9.5 1,7,7 5 1,7,12 6.67 Mean 6 1,7,12 6.67 Std error 2.45 3,7,7 5.67 3,7,12 7.33 3,7,12 7.33 2 1 Distribution, n=2 3.5 1 Mean 6 4 2 Std error 2.00 3.5 4.5 1 3 5 2 frequency 2.5 6.5 3 2 7 1 1.5 1 7.5 1 0.5 9 1 0 9.5 2 2 3.5 4 4.5 5 6.5 7 7.5 9 9.5 TOTAL 15 days absent 3.33 1 3.67 2 Distribution, n=3 4.67 2 5 1 3.5 5.33 3 3 5.67 1 2.5 frequency 6.33 1 2 6.67 3 1.5 7 1 1 7.33 2 0.5 8.33 2 0 1 2 3 4 5 6 7 8 9 10 11 12 8.67 1 days absent TOTAL 20 5. a. .3085 somewhat likely b. .383 c. [1.27, 1.33] d. 1.3 and .01 e. normal because the original population is normal f. 0.0228 very unlikely g. 0.9544 h. [1.29, 1.31] i. A sample mean. Mean of samples are much more likely to fall near the population mean. 6. µ = 150 seconds, σ = 15 seconds a. .1293 or 12.93% b. .4756 or 47.56% (notice the mean of a sample is much more likely to be in this interval than an individual observation) Page 7 of 7

DOCUMENT INFO

Shared By:

Categories:

Tags:

Stats:

views: | 13 |

posted: | 8/28/2011 |

language: | English |

pages: | 7 |

OTHER DOCS BY niusheng11

How are you planning on using Docstoc?
BUSINESS
PERSONAL

By registering with docstoc.com you agree to our
privacy policy and
terms of service, and to receive content and offer notifications.

Docstoc is the premier online destination to start and grow small businesses. It hosts the best quality and widest selection of professional documents (over 20 million) and resources including expert videos, articles and productivity tools to make every small business better.

Search or Browse for any specific document or resource you need for your business. Or explore our curated resources for Starting a Business, Growing a Business or for Professional Development.

Feel free to Contact Us with any questions you might have.